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PROFESSOR: Welcome back to 8.20.

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In this section, you want to
look at light, what is it,

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and how does it propagate.

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In this video, specifically,
I give you a little bit

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of a preview of 8.02.

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And I don't do this in
a very topological way.

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I just give you
some information.

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So if we study 8.02, we'll
see Maxwell equations

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are being developed in there.

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We look at Maxwell equations
for electric and magnetic field

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E and B in vacuum.

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We can rewrite the
Maxwell equations

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and define wave equations.

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The solutions of
the wave equation,

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as the name tells
you, are waves.

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So what we are
looking at here is

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you want to describe
the propagation

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of electric and magnetic
fields in vacuum.

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In this situation, this is maybe
at some time, t equal to 0,

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we have an electric
field in this point here,

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and a magnetic field--
electric field points

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into the y direction,
the magnetic field

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into the z direction.

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And what the
equations now describe

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is how the wave propagates
in space and in time.

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And you can already tell
from the name, wave equation,

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the solutions of this equation--

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these differential
sines and cosines.

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So one solution here
is are Ey equal to E0,

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times cosine, kx minus omega t.

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We find then that the speed
in which the wave propagates--

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you pick one peak
of a wave, and you

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see how it propagates--
one point of variance

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here propagates.

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The speed in which it propagates
is the speed of light, c.

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And you can find c here through
those constants in the Maxwell

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equations and wave equations.

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Find c is 1 over square
root epsilon 0 and mu 0.

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The permeativity and
the permeability,

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the product of the two gives
you the speed of light.

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So if you look at
this some more,

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and connect the Maxwell
equation to the Lorentz force,

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again, as a reminder, for
those who had had 8.02 already,

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the force on the
charged particle

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in an electromagnetic
field is given by 2 times

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E, plus 2 times V cross B.

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If you have two charges, the
force between those two charges

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is the product of the 2
divided by r squared, times 1

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over 4 pi x mu 0.

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Again, [INAUDIBLE].

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And the force
between two wires--

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this current-- current
i1 and current i2--

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is equal to the product of the
two currents, divided by r,

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times l-- the length
of the wires--

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times mu 0 over 2 pi.

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So this is fantastic,
because now you

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can calculate the
speed of light by just

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measuring the forces
between charges

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and current in wire's centers.

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The value of c is
also very interesting.

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It's large-- very large.

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3 times 10 to the 8
meters per second.

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So just let that sink in.

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We, as humans, move with
a few meters per second.

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Light travels-- a few
nanoseconds is needed for light

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to travel about 1 meter.

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It takes just nanosecond.

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Let's stop the video here.

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The next thing I want
to do is an exercise.

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I want to have you play with
this differential equation,

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and there's a solution of
the differential equation.

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But the challenge
or the exercise

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is to show that if you have a
function which you can write

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as f0, which is an
arbitrary function, which

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is a function of x minus
ct, those functions,

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regardless in how
they look like,

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are solutions of this
differential equation.

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Note that I replaced our
constant epsilon 0 and mu

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0 now with 1 over c squared.

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So f0 can really be
an arbitrary function.

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You need to be able to build
the derivative, though.

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So I do the function
here as a function

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of x for some time equal t0.

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And then I drew the same
function 4 times equal to 1.

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And so you can,
from this picture,

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see that the delta x over delta
t is minus c in this case.

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So my function-- my wave is
moving with the speed of light

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in minus direction.

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So I want you to show that this
kind of equation [INAUDIBLE]

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wave equation.

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So I would like you to do
this, and stop the video,

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and show you the solution next.

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So the way to approach this is
simply applying the chain rule.

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And that might be something
you want to remind yourself of.

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So after I do this, I'll define
this little helper function

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here u is equal to x minus ct.

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And this makes our
function a function

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of u, which is itself
a function of x and t.

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So if I built a
derivative with x,

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I have this df of
u, du times du dx.

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If I build a second
derivative, there

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is a product you
have to take care of.

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So I find that d is the second
derivative of f of u here,

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times du dx squared.

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And then I have to add df du
times second derivative of u.

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This follows very similar
for the derivative of t.

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And then I can
investigate what we find.

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So my du dx is equal to 1--

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du dx.

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If I build the derivative of
x minus et, this x, I find 1.

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I do the same with t--

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I find minus c.

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I will use this second
derivatives of u.

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x and t are all 0.

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If I put this now
in my equation,

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I find second derivative of
f with u is of 1 minus c--

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sorry, 1 minus 1 over c squared
times c squared is equal to 0.

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And since this is
always 0, we have just

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proven that any sort
of function which

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I can build the derivative of
which is of the from x minus ct

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solves that equation.